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Harman [31]
3 years ago
10

The sum of two numbers is 27. Their difference is 1. Find the smaller number.​

Mathematics
1 answer:
hjlf3 years ago
6 0

Answer:

12 and 15.

Hope this helped

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If f(x) = 2(x − 5), find f(8).
Dafna11 [192]

Answer:

The answer is eight.

Step-by-step explanation:

We know that x is eight. That means we can plug eight in for x. If we do that we get 2^8-5. 8-5 is three, so our problem becomes 2^3, which is 8.

4 0
3 years ago
Question is shown in the image
astraxan [27]
Your average rate of change for the interval is 3/2 or 1.5, so that should be your answer.
6 0
2 years ago
Two towers make up the Vertical Forest. One is 76 meters tall, and the other is 111 meters tall. How many feet taller is the hig
prisoha [69]

Answer: 114.8

Step-by-step explanation: first multiply 76 times 3.28 (249.28) then multiply 11 by 3.28 (364.08) and subtract to get your awnser

6 0
2 years ago
Evaluate the expression, if possible: √ 64
astraxan [27]

The square root of a number is the smallest (positive) number possible that when multiplied by it self equals the number in the square root.

The square root of 64 is 8 ( 8 x 8 = 64).

The answer is B.

7 0
3 years ago
The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?
Lina20 [59]
\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right)  (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y


y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}

7 0
3 years ago
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